INTERNATIONAL JOURNAL OF 2013 Institute for Scientiﬁc · in the viscoelastic model of Oldroyd...

INTERNATIONAL JOURNAL OF © 2013 Institute for ScientificNUMERICAL ANALYSIS AND MODELING Computing and InformationVolume 10, Number 2, Pages 481–507

ON FULLY DISCRETE FINITE ELEMENT SCHEMES FOR

EQUATIONS OF MOTION OF KELVIN-VOIGT FLUIDS

SAUMYA BAJPAI, NEELA NATARAJ AND AMIYA K. PANI

Abstract. In this paper, we study two fully discrete schemes for the equations of motion arisingin the Kelvin-Voigt model of viscoelastic fluids. Based on a backward Euler method in time and afinite element method in spatial direction, optimal error estimates which exhibit the exponentialdecay property in time are derived. In the later part of this article, a second order two stepbackward difference scheme is applied for temporal discretization and again exponential decay intime for the discrete solution is discussed. Finally, a priori error estimates are derived and resultson numerical experiments conforming theoretical results are established.

Key words. Viscoelastic fluids, Kelvin-Voigt model, a priori bounds, backward Euler method,second order backward difference scheme, optimal error estimates.

1. Introduction

In this article, we discuss the convergence of the backward Euler method andthe second order backward difference scheme for the following system of equationsof motion arising in the Kelvin-Voigt fluids (see [18]):

∂u

∂t+ u · ∇u− κ∆ut − ν∆u+∇p = f(x, t), x ∈ Ω, t > 0,(1.1)

and incompressibility condition

∇ · u = 0, x ∈ Ω, t > 0,(1.2)

with initial and boundary conditions

u(x, 0) = u0 in Ω, u = 0 on ∂Ω, t ≥ 0,(1.3)

where, Ω is a bounded domain in IRd (d = 2 or 3) with boundary ∂Ω. Hereu = u(x, t) represents the velocity vector, p = p(x, t) the pressure and ν > 0,the kinematic coefficient of viscosity. Moreover, the velocity of the fluid, after in-stantaneous removal of the stress, does not vanish instantaneously but dies outlike exp(κ−1t) (see [18]), where κ is the retardation parameter. For details of thephysical background and its mathematical modeling, we refer to [6]-[7] and [9].Throughout this paper, we assume that the right hand side function f = 0. In fact,assuming conservative force, the function f can be absorbed in the pressure term.Based on the analysis of Ladyzenskaya [16] for the solvability of the Navier Stokesequations, Oskolkov [17, 18], has proved the global existence of a unique ‘almost’classical solution in finite time interval for the initial and boundary value problem(1.1)-(1.3). The investigations on solvability are further continued by him and hiscollaborators, see [20] and [21] and they have discussed the existence and unique-ness results on the entire semiaxis R+ in time.For the related literature on the time discretization of equations of motion arisingin the viscoelastic model of Oldroyd type see [2], [12], [23] and [25]-[28]. Interest-ingly, there is hardly any work devoted to the time discretization of (1.1)-(1.3). Forthe earlier results on the numerical approximations to the solutions of the problem

Received by the editors April 19, 2012.2000 Mathematics Subject Classification. 65M60,65M12,65M15,35D05,35D10.

481

482 S. BAJPAI, N. NATARAJ AND A. PANI

(1.1)-(1.3), we refer to [3] and [19]. Under the condition that the solution is asymp-totically stable as t → ∞, the authors of [19] have established the convergence ofspectral Galerkin approximations for the semi axis t ≥ 0. Recently, Bajpai et al. [3]have applied finite element methods to discretize the spatial variables and derivedoptimal error bounds for the velocity in L∞(L2) as well as L∞(H1)-norms and forthe pressure in L∞(L2)- norm. In [3] and [19], only semidiscrete approximationsfor (1.1)-(1.3) are discussed, keeping the time variable continuous. In this article,we have discussed both backward Euler method and two step backward differencescheme for the time discretization and have derived optimal error estimates. Wehave also discussed briefly, the proof of linearized backward Euler method appliedto (1.1)-(1.3) for time discretization. More precisely, we have

‖u(tn)−Un‖j ≤ Ce−αtn(h2−j + k) j = 0, 1,

and

‖(p(tn)− Pn)‖ ≤ Ce−αtn(h+ k),

where the pair (Un, Pn) is the fully discrete solution of the backward Euler orlinearized backward Euler method.In the later part of this article, we have proved the following result for a secondorder backward difference scheme:

‖u(tn)−Un‖j ≤ Ce−αtn(h2−j + k2) j = 0, 1,

and

‖(p(tn)− Pn)‖ ≤ Ce−αtn(h+ k2−γ),

where the pair (Un, Pn) is the fully discrete solution of the second order backwarddifference scheme and

γ =

0 if n ≥ 2;

1 if n = 1.

The remaining part of this paper is organized as follows. In Section 2, we discuss thepreliminaries. In Section 3, we derive a priori bounds for the semidiscrete solutionsand present some spatial error estimates required for error analysis. In Section4, we obtain a priori bounds for the discrete solution and prove the existence anduniqueness of the discrete solution. In Section 5, we establish the error estimates forthe velocity and pressure of the backward Euler method. Section 6 deals with theerror estimates for velocity and pressure using the second order backward differencescheme. In Section 7, we provide some numerical results to confirm our theoreticalresults.

2. Preliminaries

For the mathematical formulation of (1.1)-(1.3), we denote Rd, (d = 2, 3)-valuedfunction spaces using boldface letters. That is,

H10 = (H1

0 (Ω))d, L2 = (L2(Ω))d and Hm = (Hm(Ω))d,

where L2(Ω) is the space of square integrable functions defined in Ω. The s-pace L2(Ω) is a Hilbert space endowed with the usual scalar product (φ, ψ) =∫

Ω

φ(x)ψ(x) dx and the associated norm ‖φ‖ =

(∫

Ω

|φ(x)|2 dx

)1/2

. Further, Hm(Ω)

FULLY DISCRETE SCHEMES FOR KELVIN-VOIGT FLUIDS 483

is the standard Hilbert Sobolev space of order m ∈ N+ with norm ‖φ‖m =

∑

|α|≤m

∫

Ω

|Dαφ|2 dx

1/2

. Note that H10 is equipped with a norm

‖∇v‖ =

d∑

i,j=1

(∂jvi, ∂jvi)

1/2

=

(

d∑

i=1

(∇vi,∇vi)

)1/2

.

We also use the following spaces of the vector valued functions:

J1 = φ ∈ H10 : ∇ · φ = 0,

J = φ ∈ L2 : ∇ · φ = 0 in Ω, φ · n|∂Ω = 0 holds weakly,

where n is the unit outward normal to the boundary ∂Ω and φ ·n|∂Ω = 0 should beunderstood in the sense of trace in H−1/2(∂Ω), see [24]. Let Hm/IR be the quotientspace consisting of equivalence classes of elements ofHm differing by constants, withnorm ‖p‖Hm/IR = infc∈IR ‖p+ c‖m. Let P be the orthogonal projection of L2 ontoJ.We need further assumptions, that is,(A1). For g ∈ L2, let v ∈ J1, q ∈ L2/IR be the unique pair of solution to thesteady state Stokes problem, see [24],

−∆v +∇q = g,

∇ · v = 0 in Ω, v|∂Ω = 0

satisfying the following regularity result:

‖v‖2 + ‖q‖H1/IR ≤ C‖g‖.(2.1)

Setting

−∆ = −P∆ : J1 ∩H2 ⊂ J → J

as the Stokes operator, (A1) implies

‖v‖2 ≤ C‖∆v‖ ∀v ∈ J1 ∩H2.(2.2)

It is easy to show that

‖v‖2 ≤ λ−11 ‖∇v‖2 ∀v ∈ H1

0(Ω),(2.3)

‖∇v‖2 ≤ λ−11 ‖∆v‖2 ∀v ∈ J1 ∩H2.

where λ−11 is a positive constant depending on the domain Ω. In fact, this is known

as Poincare inequality with λ−11 as the best possible positive constant.

(A2). There exists a positive constant M , such that the initial velocity u0 satisfies

u0 ∈ H2 ∩ J1 with ‖u0‖2 ≤M.

Moreover, we define a bilinear form a(·, ·) on H10 ×H1

0 by

a(v,φ) = (∇v,∇φ) ∀v, φ ∈ H10,(2.4)

and a trilinear form b(·, ·, ·) on H10 ×H1

0 ×H10 by

b(v,w,φ) =1

2(v · ∇w,φ)−

1

2(v · ∇φ,w) ∀v,w,φ ∈ H1

0.(2.5)


With the help of above notations, the variational formulation of problem (1.1)-(1.3)with f = 0 is defined as follows: Find u(t) ∈ J1 such that

(ut,φ) + κ a(ut,φ) + ν a(u,φ) + b(u,u,φ) = 0 ∀φ ∈ J1 t > 0,(2.6)

u(0) = u0.

3. Finite Element Approximation

Let h > 0 be a discretization parameter. Further, let Hh and Lh, 0 < h < 1 befinite dimensional subspaces of H1

0 and L2, respectively. Assume that the subspaceHh and Lh satisfy the following approximation properties:(B1). For each w ∈ J1 ∩H2 and q ∈ H1/IR, there exist approximations ihw ∈ Jh

and jhq ∈ Lh such that

‖w − ihw‖ + h‖∇(w− ihw)‖ ≤ K0h2‖w‖2, ‖q − jhq‖L2/IR ≤ K0h‖q‖H1/IR.

We define the subspace Jh of Hh as follows:

Jh = vh ∈ Hh : (χh,∇ · vh) = 0 ∀χh ∈ Lh.

Note that, the space Jh is not a subspace of J1. The discrete analogue of theweak formulation (2.6) is as follows: find uh(t) ∈ Hh and ph(t) ∈ Lh such thatuh(0) = u0h and for t > 0,

(uht,φh) + κ a(uht,φh) + ν a(uh,φh) + b(uh,uh,φh)

− (ph,∇ · φh) = 0 ∀φh ∈ Hh,(3.1)

(∇ · uh, χh) = 0 ∀χh ∈ Lh.

Equivalently, find uh(t) ∈ Jh such that uh(0) = u0h and for t > 0,

(uht,φh) + κ a(uht,φh) + ν a(uh,φh) = −b(uh,uh,φh) ∀φh ∈ Jh.(3.2)

Once we compute uh(t) ∈ Jh, the approximation ph(t) ∈ Lh to the pressure p(t)can be found out by solving the following system

(ph,∇ · φh) = (uht,φh) + κ a(uht,φh) + ν a(uh,φh)

+ b(uh,uh,φh) ∀φh ∈ Hh.(3.3)

For solvability of the systems (3.2) and (3.3), see [3]. Uniqueness is obtained in thequotient space Lh/Nh, where

Nh = qh ∈ Lh : (qh,∇ · φh) = 0 ∀φh ∈ Hh.

The norm on Lh/Nh is given by

‖qh‖L2/Nh= inf

χh∈Nh

‖qh + χh‖.

Furthermore, the pair (Hh, Lh/Nh) satisfies a uniform inf-sup condition:(B2). For every qh ∈ Lh, there exist a non-trivial function φh ∈ Hh and a positiveconstant K1, independent of h, such that,

|(qh,∇ · φh)| ≥ K1‖∇φh‖‖qh‖L2/Nh.

As a consequence of conditions (B1), we have the following properties of the L2

projection Ph : L2 → Jh. For φ ∈ J1, we note that, see ([11], [13]),

‖φ− Phφ‖+ h‖∇Phφ‖ ≤ Ch‖∇φ‖,(3.4)

and for φ ∈ J1 ∩H2,

‖φ− Phφ‖+ h‖∇(φ− Phφ)‖ ≤ Ch2‖∆φ‖.(3.5)


We now define the discrete operator ∆h : Hh → Hh through the bilinear forma(·, ·) as

a(vh,φh) = (−∆hvh,φh) ∀vh,φh ∈ Hh.(3.6)

Set the discrete analogue of the Stokes operator ∆ = P∆ as ∆h = Ph∆h. UsingSobolev embedding theorems with Sobolev inequalities, it is a routine calculationto derive the following lemma, see page 360 of [14].

Lemma 3.1. The trilinear form b(·, ·, ·) satisfies the following estimates for allφ, ξ, χ ∈ Hh:

|b(φ, ξ, χ)| ≤ C‖∇φ‖1/2‖∆hφ‖1/2‖∇ξ‖ ‖χ‖,(3.7)

|b(φ, ξ, χ)| ≤ C‖∇φ‖‖∇ξ‖1/2‖∆hξ‖1/2‖χ‖,(3.8)

|b(φ, ξ, χ)| ≤ C‖φ‖12 ‖∇φ‖

12 ‖∇ξ‖‖∇χ‖.(3.9)

Note that, the operator b(·, ·, ·) preserves the antisymmetric properties of the orig-inal nonlinear term, that is,

b(vh,wh,wh) = 0 ∀vh,wh ∈ Hh.(3.10)

Examples of subspaces Hh satisfying assumptions (B1) and (B2) can be found in[4], [5] and [13].Below, we derive some a priori estimates for the discrete solution uh of (3.2) anal-ogous to those known for continuous solution u of (2.6) (see [3]).

Lemma 3.2. Let 0 ≤ α <νλ1

2(1 + κλ1)and u0h = Phu0, and the assumptions (A1)–

(A2) hold true. Then, the solution uh of (3.2) satisfies

‖uh(t)‖2 + κ‖∇uh(t)‖

2 + κ‖∆huh(t)‖2

+ βe−2αt

∫ t

0

e2αs(‖∇uh(s)‖2 + ‖∆huh(s)‖

2) ds ≤ C(κ, ν, α, λ1,M)e−2αt t > 0,

where β = ν − 2α(λ−11 + κ) > 0.

Proof. Setting uh(t) = eαtuh(t) for some α ≥ 0, we rewrite (3.2) as

(uht,φh)−α(uh,φh) + κ(∇uht,∇φh)− κα(∇uh,∇φh)(3.11)

+ ν(∇uh,∇φh) + e−αtb(uh, uh,φh) = 0 ∀φh ∈ Jh.

Choose φh = uh in (3.11). Using (3.10), b(uh, uh, uh) = 0 and from (2.3), we findthat

d

dt(‖uh‖

2 + κ‖∇uh‖2) + 2

(

ν − α(

κ+1

λ1))

‖∇uh‖2 ≤ 0.(3.12)

Integrate (3.12) with respect to time from 0 to t to obtain

‖uh‖2 + κ‖∇uh‖

2 + 2βe−2αt

∫ t

0

e2αs‖∇uh(s)‖2ds

≤ e−2αt(‖u0h‖2 + κ‖∇u0h‖

2).(3.13)

Using the discrete Stokes operator ∆h, we rewrite (3.11) as

(uht,φh)− α(uh,φh)− κ(∆huht,φh) + κα(∆huh,φh)(3.14)

− ν(∆huh,φh) = −e−αtb(uh, uh,φh).


We note that −(uht, ∆huh) =12

ddt‖∇uh‖

2. With φh = −∆huh, (3.14) becomes

d

dt(‖∇uh‖

2 + κ‖∆huh‖2) + 2(ν − κα)‖∆huh‖

2(3.15)

= 2α‖∇uh‖2 + 2e−αtb(uh, uh, ∆huh).

To estimate the nonlinear term on the right hand side of (3.15), a use of (3.7) yields

|I| = |e−αtb(uh, uh, ∆huh)| ≤ C‖∇uh‖32 ‖∆huh‖

32 .(3.16)

Applying Young’s inequality ab ≤ ap

pǫp/q+ ǫbq

q , a, b ≥ 0, ǫ > 0 with p = 4 and q = 43 ,

we obtain

|I| ≤ C‖∇uh‖

6

4ǫ3+

3ǫ

4‖∆huh‖

2.(3.17)

Choosing ǫ = 4ν3 , we find that

|I| ≤C

4

(

3

4ν

)3

‖∇uh‖6 + ν‖∆huh‖

2.(3.18)

Substitute (3.18) in (3.15) to arrive at

d

dt(‖∇uh‖

2 + κ‖∆huh‖2) + (ν − 2ακ)‖∆huh‖

2

≤ C(ν)‖∇uh‖6 + 2α‖∇uh‖

2.(3.19)

An integration of (3.19) with respect to time from 0 to t yields

‖∇uh‖2 + κ‖∆huh‖

2 + β

∫ t

0

‖∆huh(s)‖2ds ≤ ‖∇u0h‖

2(3.20)

+ κ‖∆hu0h‖2 + C(ν, α)

∫ t

0

(

‖∇uh(s)‖6ds+ ‖∇uh(s)‖

2)

ds.

Using (3.13), we bound

∫ t

0

‖∇uh(s)‖6ds =

∫ t

0

‖∇uh(s)‖4‖∇uh(s)‖

2ds

≤ C(κ)(‖u0h‖2 + κ‖∇u0h‖

2)2∫ t

0

‖∇uh(s)‖2ds

≤ C(κ, ν, α, λ1)(‖u0h‖2 + κ‖∇u0h‖

2)3.(3.21)

Substitute (3.21) and (3.13) in (3.20) and use stability properties of Ph to obtain

‖∇uh‖2 + κ‖∆huh‖

2 + βe−2αt

∫ t

0

e2αs‖∆huh(s)‖2ds ≤

(

‖∇u0h‖2

+ κ‖∆hu0h‖2 + C(κ, ν, α, λ1)

(

‖u0h‖2 + κ‖∇u0h‖

2)3

(3.22)

+ C(κ, ν, α, λ1)(

‖u0h‖2 + κ‖∇u0h‖

2)

)

e−2αt

≤ C(κ, ν, α, λ1,M)e−2αt.

Combine (3.13) with (3.22) to complete the rest of the proof.

In the following three lemmas, we derive a priori estimates involving time deriva-tives of the semi-discrete solution.



2(1 + κλ1)and let the assumptions (A1)–(A2) hold

true. Then, there is a positive constant C = C(κ, ν, α, λ1,M), such that for allt > 0,

‖uht(t)‖2 + κ‖∇uht(t)‖

2 + e−2αt

∫ t

0

e2αs(‖uht(s)‖2 + κ‖∇uht(s)‖

2)ds ≤ Ce−2αt.

Proof. Substituting φh = uht in (3.2), we obtain

‖uht‖2 + κ‖∇uht‖

2 = −ν(∇uh,∇uht)− b(uh,uh,uht)

= I1 + I2, say.(3.23)

To estimate |I1|, we apply Cauchy-Schwarz’s inequality and Young’s inequality toarrive at

|I1| ≤ν

2ǫ‖∇uh‖

2 +ǫ

2‖∇uht‖

2.(3.24)

Choose ǫ = κ in (3.24) to obtain

|I1| ≤ C(ν, κ)‖∇uh‖2 +

κ

2‖∇uht‖

2.(3.25)

An application of (3.7) and Young’s inequality yields

|I2| ≤ C‖∇uh‖12 ‖∆huh‖

12 ‖∇uh‖‖uht‖

≤ C‖∇uh‖3‖∆huh‖+

1

2‖uht‖

2.(3.26)

A use of (3.25), (3.26) and Lemma 3.2 in (3.23) yields

‖uht‖2 + κ‖∇uht‖

2 ≤ C(κ, ν, α, λ1,M)e−2αt.(3.27)

Next, substituting φh = e2αtuht in (3.2), we arrive at

e2αt(‖uht‖2 + κ‖∇uht‖

2) = −νe2αta(uh,uht)− e2αtb(uh,uh,uht).(3.28)

Using Cauchy-Schwarz’s inequality, (3.9), (2.3), Young’s inequality and integratingfrom 0 to t with respect to time, we obtain

∫ t

0


2)ds ≤ C(κ, ν, λ1)(

∫ t

0

e2αs(‖∇uh(s)‖2

+ ‖∇uh(s)‖4)ds

)

.(3.29)

A use of Lemma 3.2 to bound∫ t

0

e2αs‖∇uh(s)‖4ds ≤ C(κ, ν, α, λ1,M)e−2αt

∫ t

0

e2αs‖∇uh(s)‖2ds

≤ C(κ, ν, α, λ1,M)e−2αt.(3.30)

An application of (3.30) and Lemma 3.2 in (3.29) yields∫ t

0


2)ds ≤ C(κ, ν, α, λ1,M).(3.31)

A combination of (3.27) and (3.31) would lead us to the desired result.



true. Then, there is a positive constant C = C(κ, ν, α, λ1,M) such that for allt > 0,

‖uhtt(t)‖2 + κ‖∇uhtt(t)‖

2 + e−2αt

∫ t

0

e2αs(‖uhtt(s)‖2 + κ‖∇uhtt(s)‖

2)ds ≤ Ce−2αt.


Proof. Differentiation of (3.2) with respect to time yields

(uhtt,φh) + κa(uhtt,φh) + νa(uht,φh) + b(uht,uh,φh)(3.32)

+ b(uh,uht,φh) = 0 ∀φh ∈ Jh t > 0.

Substitute φh = uhtt in (3.32) to obtain

‖uhtt‖2 + κ‖∇uhtt‖

2 = −νa(uht,uhtt)− b(uht,uh,uhtt)

− b(uh,uht,uhtt).(3.33)

An application of Cauchy-Schwarz’s inequality, Young’s inequality, (3.9) and (2.3)yields

‖uhtt‖2 + κ‖∇uhtt‖

2 ≤ C(κ, ν, λ1)(

‖∇uht‖2

+ ‖∇uh‖2‖∇uht‖

2)

.(3.34)

With the help of estimates obtained from Lemma 3.2 and 3.3, we write

‖uhtt(t)‖2 + κ‖∇uhtt(t)‖

2 ≤ C(κ, ν, α, λ1,M)e−2αt.(3.35)

Multiply (3.34) by e2αt and integrate with respect to time from 0 to t to arrive at∫ t

0


2)ds ≤ C(κ, ν, λ1)

(∫ t

0

e2αs(

‖∇uht(s)‖2

+ ‖∇uh(s)‖2‖∇uht(s)‖

2)

ds

)

.(3.36)

Applying the estimates from Lemmas 3.2 and 3.3, we obtain the desired result, thatis,

∫ t

0


2)ds ≤ C(κ, ν, α, λ1,M).(3.37)

A use of (3.35) and (3.37) completes the proof.

Differentiating (3.32) with respect to time and proceeding as in the proofs of Lem-mas 3.3 and 3.4, we arrive at following Lemma:



true. Then, there is a positive constant C = C(κ, ν, α, λ1,M), such that for allt > 0,

‖uhttt(t)‖2 + κ‖∇uhttt(t)‖

2 + e−2αt

∫ t

0

e2αs(‖uhttt(s)‖2 + κ‖∇uhttt(s)‖

2)ds ≤ Ce−2αt.

Before proceeding to the error analysis for time discretization, we recall the follow-ing bounds of the error (u− uh, p− ph) (for a proof see [3]):

Theorem 3.1. Let assumptions (A1)-(A2) and (B1)-(B2) be satisfied and letu0h = Phu0. Then, there exists a positive constant C depending on λ1, κ, ν, α

and M , such that, for all t > 0 and for 0 ≤ α <νλ1

2(

1 + λ1κ) , the following estimate

holds true :

‖(u− uh)(t)‖ + h‖∇(u− uh)(t)‖+ h‖(p− ph)(t)‖L2/Nh≤ Ch2e−αt.

Remark. For similar semidiscrete error estimates of the viscoelastic model ofOldroyd type, we refer [22].


4. Backward Euler Method

In this section, we consider a backward Euler method for time discretization ofthe finite element Galerkin approximation (3.1). Let tn

Nn=0 be a uniform parti-

tion of [0, T ], and tn = nk, with time step k > 0. For smooth function φ defined

on [0, T ], set φn = φ(tn) and ∂tφn =

(φn−φn−1)

k .Now, the backward Euler method applied to (3.1) determines a sequence of func-tions Unn≥1 ∈ Hh and Pnn≥1 ∈ Lh as solutions of the following recursivenonlinear algebraic equations:

(∂tUn,φh) + κa(∂tU

n,φh) + νa(Un,φh)

+ b(Un,Un,φh) = (Pn,∇ · φh) ∀φh ∈ Hh,(4.1)

(∇ ·Un, χh) = 0 ∀χh ∈ Lh,

U0 = u0h.

Equivalently, for φh ∈ Jh, we seek Unn≥1 ∈ Jh such that


n,φh) + νa(Un,φh) + b(Un,Un,φh) = 0 ∀φh ∈ Jh,(4.2)

U0 = u0h.

Now, to study the issue of the existence and uniqueness of the discrete solutionsUnn≥1, we derive a priori bounds for the solution Unn≥1.

Lemma 4.1. With 0 ≤ α <νλ1

2(1 + λ1κ), choose k0 so that for 0 < k ≤ k0

νkλ1κλ1 + 1

+ 1 > eαk.(4.3)

Then the discrete solution UN , N ≥ 1 of (4.2) satisfies

(‖UN‖2 + κ‖∇UN‖2) + 2β1e−2αtN k

N∑

n=1

e2αtn‖∇Un‖2 ≤ e−2αtN (‖U0‖2 + κ‖∇U0‖2),

where

β1 =

(

e−αkν − 2(1− e−αk

k

)(

κ+1

λ1

)

)

> 0.(4.4)

Proof. Multiplying (4.2) by eαtn and setting Un = eαtnUn, we obtain

eαtn(


n,φh)

)

+ νa(Un,φh)

+ e−αtnb(Un, Un,φh) = 0 ∀φh ∈ Jh.(4.5)

Note that,

eαtn ∂tUn = eαk∂tU

n −

(

eαk − 1

k

)

Un.(4.6)

Using (4.6) in (4.5) and multiplying the resulting equation by e−αk, we obtain


n,φh)−

(

1− e−αk

k

)

(Un,φh) + e−αkνa(Un,φh)(4.7)

− κ

(

1− e−αk

k

)

a(Un,φh) + e−αtn+1b(Un, Un,φh) = 0.


Note that

(∂tUn, Un) ≥

1

2k(‖Un‖2 − ‖Un−1‖2) =

1

2∂t‖U

n‖2.(4.8)

Substituting φh = Un in (4.7) and using (2.3) along with (3.10) yields

1

2∂t(‖U

n‖2 + κ‖∇Un‖2) +

(

e−αkν −(1− e−αk

k

)(

κ+1

λ1

)

)

‖∇Un‖2 ≤ 0.(4.9)

Note that, the coefficient of the second term on the left hand side is greater than

β1. With 0 ≤ α <νλ1

2(1 + λ1κ), choose k0 > 0 such that for 0 < k ≤ k0

νkλ11 + κλ1

+ 1 > eαk.

Then, for 0 < k ≤ k0, the coefficient β1 (see (4.4)) of the second term on the lefthand side of (4.9) becomes positive. Multiplying (4.9) by 2k and summing overn = 1 to N , we obtain

‖UN‖2 + κ‖∇UN‖2 + 2β1k

N∑

n=1

‖∇Un‖2 ≤ ‖U0‖2 + κ‖∇U0‖2.(4.10)

Divide (4.10) by e2αtN to complete the rest of the proof.

Theorem 4.1. (Brouwer’s fixed point theorem)[15]. Let H be a finite dimensionalHilbert space with inner product (·, ·) and ‖ · ‖. Let g : H → H be a continuousfunction. If there exists R > 0 such that (g(z), z) > 0 ∀z with ‖z‖ = R, then thereexists z∗ ∈ H such that ‖z‖ ≤ R and g(z∗) = 0.

Now, we are ready to prove the following existence and uniqueness result.

Theorem 4.2. Given Un−1, the discrete problem (4.2) has a unique solution Un,n ≥ 1.

Proof. Given Un−1, define a function F : Jh → Jh for a fixed ′n′ by

(F(v),φh) = (v,φh) + κ(∇v,∇φh) + kν(∇v,∇φh)(4.11)

+ k b(v,v,φh)− (Un−1,φh)− κ(∇Un−1,∇φh).

Define a norm on Jh as

‖|v‖| = (‖v‖2 + κ‖∇v‖2)12 .(4.12)

We can easily show that F is continuous. Now, after substituting φh = v in (4.11),we use (3.10), (4.12), Cauchy-Schwarz’s inequality and Young’s inequality to arriveat

(F(v),v) ≥ (‖|v‖| − ‖|Un−1‖|)‖|v‖|.

Choose R such that ‖|v‖| = R and R− ‖|Un−1‖| > 0 and hence,

(F(v),v) > 0.

A use of Theorem 4.1 would provide us the existence of Unn≥1.Now, to prove uniqueness, set En = Un

1 −Un2 , where Un

1 and Un2 are the solutions

of (4.2).Note that,

(∂tEn,φh) + κa(∂tE

n,φh) + νa(En,φh)

= b(Un2 ,U

n2 ,φh)− b(Un

1 ,Un1 ,φh).(4.13)


Using φh = En and proceeding as in the derivation of (4.9), we obtain

1

2∂t(‖E

n‖2 + κ‖∇En‖2) +

(


k

)(

κ+1

λ1

)

)

‖∇En‖2(4.14)

≤ e−αkeαtnΛn1 (E

n),

where

Λn1 (E

n) = −b(Un1 ,U

n1 , E

n) + b(Un2 ,U

n2 , E

n).

Note that,

eαtnΛn1 (E

n) = e−αtn |b(Un1 , U

n1 , E

n)− b(Un2 , U

n2 , E

n)|(4.15)

= e−αtn |b(En, Un1 , E

n) + b(Un2 , E

n, En)|.

A use of (3.10), (3.9) and (2.3) in (4.15) yields

eαtn |Λn1 (E

n)| ≤ Ce−αtn‖En‖12 ‖∇En‖

12 ‖∇Un

1‖‖∇En‖(4.16)

≤ C(λ1)e−αtn‖∇Un

1 ‖‖∇En‖2.

Using (4.16), E0 = 0, Young’s inequality in (4.14), multiplying by 2k, summingover n = 1 to N and applying the bounds of Lemma 4.1, we arrive at

‖EN‖2 + κ‖∇EN‖2 ≤ C(ν, λ1)ke−αk

N−1∑

n=1

e−2αtn‖∇Un1 ‖

2‖∇En‖2

+ C(ν, λ1)ke−αke−2αtN‖∇UN

1 ‖2‖∇EN‖2

≤ C(ν, λ1)ke−αk

N−1∑

n=0

e−2αtn‖∇Un1 ‖

2‖∇En‖2

+ C(ν, λ1, κ,M)ke−αk(‖EN‖2 + κ‖∇EN‖2).(4.17)

Since, (1 − C(ν, λ1, κ,M)ke−αk) can be made positive for small k, an applicationof the discrete Gronwall’s Lemma and Lemma 4.1 in (4.17) yields

‖EN‖2 + κ‖∇EN‖2 ≤ 0(4.18)

and this provides the uniqueness of the solutions Unn≥1.

5. Error Analysis for Backward Euler Method

In this section, we obtain the H1 and L2- norm estimates for the error en =Un−uh(tn) = Un−un

h and the L2- norm estimate for the error ρn = Pn−ph(tn) =Pn − pnh. The following theorem provides a bound on the error en:

Theorem 5.1. Let 0 ≤ α <νλ1

2(1 + κλ1)and k0 > 0 be such that for 0 < k ≤ k0,

(4.3) is satisfied. For some fixed h > 0, let uh(t) satisfy (3.2). Then, there existsa positive constant C, independent of k, such that for n = 1, 2, · · · , N

‖en‖2 + κ‖∇en‖2 + β1ke−2αtn

n∑

i=1

e2αti‖∇ei‖2 ≤ Ck2e−2αtn(5.1)

and

‖∂ten‖2 + ‖∂t∇en‖2 ≤ Ck2e−2αtn .(5.2)


Proof. Consider (3.2) at t = tn and subtract it from (4.2) to obtain

(∂ten,φh) + κa(∂te

n,φh) + νa(en,φh)(5.3)

= (σn1 ,φh) + κa(σn

1 ,φh) + Λh(φh) ∀φh ∈ Jh,

where σn1 = un

ht − ∂tunh and Λh(φh) = b(un

h,unh,φh)− b(Un,Un,φh). Multiplying

(5.3) by eαtn , we arrive at

(eαtn ∂ten,φh) + κa(eαtn ∂te

n,φh) + νa(en,φh)(5.4)

= (eαtnσn1 ,φh) + κa(eαtnσn

1 ,φh) + eαtnΛh(φh).

Note that,

eαtn ∂ten = eαk∂te

n −(eαk − 1

k

)

en.(5.5)

Using (5.5) in (5.4) and dividing the resulting equation by eαk, we obtain


n,φh)− (1 − e−αk

k)(en,φh)(5.6)

− (1 − e−αk

k)κa(en,φh) + νe−αka(en,φh) = e−αk(eαtnσn

1 ,φh)

+ e−αkκa(eαtnσn1 ,φh) + e−αkeαtnΛh(φh).

Substitute φh = en in (5.6). A use of (2.3) yields

1

2∂t(

‖en‖2 + κ‖∇en‖2)

+

(

νe−αk −(1− e−αk

k

)(

κ+1

λ1

)

)

‖∇en‖2(5.7)

= e−αk(eαtnσn1 , e

n) + e−αkκa(eαtnσn1 , e

n) + e−αkeαtnΛh(en).

On multiplying (5.7) by 2k and summing over n = 1 to N , we observe that

‖eN‖2 + κ‖∇eN‖2 + 2k

(


k

)(

κ+1

λ1

)

) N∑

n=1

‖∇en‖2(5.8)

≤ 2ke−αkN∑

n=1

(eαtnσn1 , e

n) + 2ke−αkN∑

n=1

κa(eαtnσn1 , e

n)

+ 2ke−αkN∑

n=1

eαtnΛh(en) = IN1 + IN2 + IN3 , say.

Using Cauchy-Schwarz’s inequality, (2.3) and Young’s inequality, we estimate IN1as:

|IN1 | ≤ 2ke−αkN∑

n=1

‖eαtnσn1 ‖‖e

n‖


N∑

n=1

‖eαtnσn1 ‖

2 +ν

3ke−αk

N∑

n=1

‖∇en‖2.(5.9)

Now, using the Taylor series expansion of uh around tn in the interval (tn−1, tn),we observe that

‖eαtnσn1 ‖

2 ≤ e2αtn1

k2

(∫ tn

tn−1

(tn − s)‖uhtt(s)‖ds

)2

.(5.10)


An application of Cauchy-Schwarz’s inequality in (5.10) yields

‖eαtnσn1 ‖

2 ≤1

k2

(∫ tn

tn−1

e2αtn‖uhtt(s)‖2ds

)(∫ tn

tn−1

(tn − s)2ds

)

=k

3

∫ tn

tn−1

e2αtn‖uhtt(t)‖2 dt(5.11)

and hence, using (5.11) and Lemma 3.4, we write

k

N∑

n=1

‖eαtnσn1 ‖

2 ≤k2

3

N∑

n=1

∫ tn

tn−1

e2αtn‖uhtt(s)‖2 ds

=k2

3e2αk

N∑

n=1

∫ tn

tn−1

e2αtn−1‖uhtt(s)‖2 ds

≤k2

3e2αk

N∑

n=1

∫ tn

tn−1

e2αs‖uhtt(s)‖2 ds

=k2

3e2αk

∫ tN

0

e2αs‖uhtt(s)‖2 ds

≤ C(κ, ν, α, λ1,M)k2e−2αtN−1 .(5.12)

Similarly, we obtain

k

N∑

n=1

‖eαtn∇σn1 ‖

2 ≤ C(κ, ν, α, λ1,M)k2e−2αtN−1.(5.13)

Using (5.12) in (5.9), we find that

|IN1 | ≤ C(κ, ν, α, λ1,M)k2 +ν

3ke−αk

N∑

n=1

‖∇en‖2.(5.14)

Following the similar steps as for bounding |IN1 | and using (5.13), we obtain

|IN2 | ≤ C(κ, ν, α, λ1,M)k2 +ν

3ke−αk

N∑

n=1

‖∇en‖2.(5.15)

To estimate IN3 , we note that

Λh(φh) = b(unh,u

nh,φh)− b(Un,Un,φh)

= b(unh,u

nh,φh)− b(Un − un

h ,Un,φh)− b(un

h,Un,φh)

= −b(unh, e

n,φh)− b(en,Un,φh).(5.16)

Hence, we find that

eαtn |Λh(en)| = e−αtn

∣

∣− b(en, Un, en)∣

∣.(5.17)

The first term of (5.16) vanish because of (3.10). A use of the generalized Holder’sinequality and Sobolev’s embedding theorems in (5.17) yields

eαtn |Λh(en)| ≤ Ce−αtn‖en‖

12 ‖∇en‖

12 ‖∇Un‖‖en‖

12 ‖∇en‖

12

≤ Ce−αtn‖∇Un‖‖en‖‖∇en‖.(5.18)


Using Young’s inequality, we arrive at

|IN3 | ≤C(ν)N−1∑

n=1

ke−αke−2αtn‖∇Un‖2 ‖en‖2 + C(ν)ke−αke−2αtN‖∇UN‖2 ‖eN‖2

+ν

3ke−αk

N∑

n=1

‖∇en‖2.(5.19)

An application of Lemma 4.1 to estimate the second term on the right hand sideof (5.19) yields

|IN3 | ≤ C(ν)

N−1∑

n=1

ke−αke−2αtn‖∇Un‖2 ‖en‖2 + C(ν,M)ke−αke−2αtN ‖eN‖2

+ν

3ke−αk

N∑

n=1

‖∇en‖2.(5.20)

A use of (5.14), (5.15) and (5.20) in (5.8) with e0 = 0 yields

‖eN‖2 + κ‖∇eN‖2 +β1kN∑

n=1

‖∇en‖2 ≤ C(κ, ν, α, λ1,M)k2

+C(ν)ke−αk

N−1∑

n=0

e−2αtn‖∇Un‖2‖en‖2

+C(ν,M)ke−αk(‖eN‖2 + κ‖∇eN‖2).(5.21)

Now choose k0 > 0 such that for 0 < k < k0, (1− C(ν,M)ke−αk) > 0 and (4.3) issatisfied. Then, an application of the discrete Gronwall’s Lemma yields

‖eN‖2 + κ‖∇eN‖2 + β1k

N∑

n=1

‖∇en‖2 ≤ C(κ, ν, α, λ1,M)k2×(5.22)

exp

(

k

N−1∑

n=0

‖∇Un‖2)

.

With the help of Lemma 4.1, we bound

kN−1∑

n=0

‖∇Un‖2 ≤ C(κ, ν, α, λ1,M).(5.23)

Using (5.23) in (5.22), we arrive at

‖eN‖2 + κ‖∇eN‖2 + β1k

N∑

n=1

‖∇en‖2 ≤ C(κ, ν, α, λ1,M)k2.(5.24)

For 0 < k ≤ k0, the coefficient of the third term on the left-hand side of (5.24),becomes positive. Dividing (5.24) by e2αtN , we obtain (5.1).Next, we take φh = ∂te

n in (5.6) and obtain

‖∂ten‖2 + κ‖∂t∇en‖2 =

(1− e−αk

k

)

(en, ∂ten)(5.25)

−(

νe−αk − κ(1− e−αk

k

))

a(en, ∂ten) + e−αk(eαtnσn

1 , ∂ten)

+ e−αkκ a(eαtnσn1 , ∂te

n) + e−αkeαtnΛh(∂ten).


Using (5.16), (3.7), (3.9) and (2.3), we observe that

eαtn |Λh(φh)| = e−αtn |b(unh, e

n,φh) + b(en, Un,φh)|

≤ C(λ1)e−αtn

(

‖∇unh‖

12 ‖∆hu

nh‖

12 + ‖∇Un‖

)

‖∇en‖‖∇φh‖.

With the help of Lemmas 3.2 and 4.1, we bound

eαtn |Λh(φh)| ≤ C(κ, ν, α, λ1,M)‖∇en‖‖∇φh‖.(5.26)

A use of Cauchy-Schwarz’s inequality, Young’s inequality, (2.3) and (5.26) in (5.25)yields

‖∂ten‖2 + κ‖∂t∇en‖2 ≤ C(κ, α, ν, λ1,M)

(

‖∇en‖2 + ‖eαtn∇σn1 ‖

2

)

.(5.27)

To estimate the second term in the right hand side of (5.27), we note from (5.11)and Lemma 3.4 that

‖eαtn∇σn1 ‖

2 ≤k

3

∫ tn

tn−1

e2αtn‖∇uhtt(t)‖2 dt

≤ C(κ, ν, α, λ1,M)ke2αtn∫ tn

tn−1

e−2αsds

≤ C(κ, ν, α, λ1,M)k2e2αk∗

,(5.28)

for k∗ ∈ (0, k). In view of (5.1) and (5.28), (5.27) implies (5.2). This completes therest of the proof. .It remains to prove the error estimate for the pressure Pn. Consider (3.1) at t = tnand subtract it from (4.1) to obtain

(ρn,∇ · φh) = (∂ten,φh) + κa(∂te

n,φh) + νa(en,φh)

− (σn1 ,φh)− κa(σn

1 ,φh)− Λh(φh).

Using Cauchy-Schwarz’s inequality, (2.3) and (5.26), we obtain

(ρn,∇ · φh) ≤ C(κ, ν, λ1)(

‖∂t∇en‖+ ‖∇en‖+ ‖∇σn1 ‖)

‖∇φh‖.(5.29)

A use of Theorem 5.1 and (5.28) in (5.29) would lead us to the desired result, thatis

‖ρn‖ ≤ C(κ, ν, α, λ1,M)k e−αtn .(5.30)

Remark 1. Note that in the estimate of IN3 , that is, the estimate (5.20), we have

used Lemma 4.1 to bound only ‖UN‖ for the second term on the right hand side of

(5.20). But we could have bounded ‖Un‖, n = 1, · · · , N−1 using again Lemma 4.1,but that would have resulted in exponential dependence of CT in the final estimate.

The backward Euler method applied to (3.1) gives rise to a nonlinear systemat t = tn. Here, we introduce a linearized version of this method which solves asystem of linear equations at each time step.

The linearized backward Euler method is as follows: find a sequence of functionsUnn≥1 ∈ Hh and Pnn≥1 ∈ Lh as solutions of the following recursive linear


algebraic equations:



+b(Un−1,Un,φh) = (Pn,∇ · φh) ∀φh ∈ Hh,(5.31)

(∇ ·Un, χh) = 0 ∀χh ∈ Lh,

U0 = u0h.

Equivalently, for φh ∈ Jh, we seek Unn≥1 ∈ Jh such that



+b(Un−1,Un,φh) = 0 ∀φh ∈ Jh,(5.32)

U0 = u0h.

The proof for the linearized backward Euler method proceeds along the same linesas in the derivation of Theorem 5.1. Here, the equation in error en is:


n,φh) + νa(en,φh) = (σn1 ,φh)(5.33)

+ κa(σn1 ,φh) + Λh(φh) ∀φh ∈ Jh,

where σn1 = un

ht − ∂tunh and Λh(φh) = b(un

h,unh,φh)− b(Un−1,Un,φh). Note that,

the difference here is only in the nonlinear term.Again, with the help of similar applications as in the proof of Theorem 5.1, wearrive at

‖eN‖2 + κ‖∇eN‖2 + 2

(


k

)(

κ+1

λ1

)

)

k

N∑

n=1

‖∇en‖2

≤ 2ke−αkN∑

n=1

(eαtnσn1 , e

n) + 2ke−αkN∑

n=1

κa(eαtnσn1 , e

n)

+ 2ke−αkN∑

n=1

eαtnΛh(en) = IN1 + IN2 + IN3 , say.(5.34)

The first two terms in the right hand side of (5.34) are bounded by (5.14) and(5.15). Hence, we need to estimate the third term. In this case, we write

eαtn |Λh(φh)| = eαtn |b(unh,u

nh,φh)− b(Un−1 − un−1

h ,Un,φh)

− b(un−1h ,Un,φh)|

= eαtn |b(unh − un−1

h ,unh,φh)− b(en−1,Un,φh)

+ b(un−1h ,un

h −Un,φh)|

= eαtn | − b(unh − un−1

h , en,φh) + b(unh − un−1

h ,Un,φh)

− b(en−1,Un,φh)− b(un−1h , en,φh)|.(5.35)

A use of (3.10) along with (2.3) and (3.9) in (5.35) with φh = en yields

eαtn |Λh(en)| ≤ eαtn |b(un

h − un−1h ,Un, en)− b(en−1,Un, en)|

≤ C(λ1)eαtn(

‖∇(unh − un−1

h )‖‖∇Un‖‖∇en‖

+ ‖∇en−1‖‖∇Un‖‖∇en‖)

.(5.36)


Hence, we observe that

|IN3 | ≤ 2ke−αk

N∑

n=1

eαtn |Λh(en)| ≤ C(λ1)ke

−αkN∑

n=1

(

eαtn‖∇(unh − un−1

h )‖ ×

‖∇Un‖‖∇en‖+ eαtn‖∇Un‖‖∇en−1‖‖∇en‖)

= |IN4 |+ |IN5 |, say.(5.37)

Note that, a use of Taylor’s series expansion of uh(t) at tn in the interval (tn−1, tn)yields

‖∇(unh − un−1

h )‖ = ‖

∫ tn

tn−1

∇uht(s)ds‖.(5.38)

With the help of Lemma 3.3 and mean value theorem, we observe that

‖∇(unh − un−1

h )‖ ≤

∫ tn

tn−1

‖∇uht(s)‖ ds ≤ C(κ, ν, α, λ1,M)

∫ tn

tn−1

e−αsds

≤ C(κ, ν, α, λ1,M)e−αtn1

α

(

eαk − 1)

≤ C(κ, ν, α, λ1,M)keαk∗

,(5.39)

where k∗ ∈ (0, k).Using Young’s inequality, (5.39) and Lemma 4.1, we bound |IN4 | as

|IN4 | ≤ C(λ1)ke−αk

N∑

n=1

e2αtn‖∇Un‖2‖∇(unh − un−1

h )‖2 +ν

6ke−αk

N∑

n=1

‖∇en‖2

≤ C(κ, ν, α, λ1,M)k2(

k

N∑

n=1

e2αtn‖∇Un‖2)

+ν

6ke−αk

N∑

n=1

‖∇en‖2

≤ C(κ, ν, α, λ1,M)k2 +ν

6ke−αk

N∑

n=1

‖∇en‖2.(5.40)

A use of Young’s inequality yields

|IN5 | = C(λ1)ke−αk

N∑

n=1

eαtn‖∇Un‖‖∇en−1‖‖∇en‖


N∑

n=1

e−2αtn−1‖∇Un‖2‖∇en−1‖2 +ν

6ke−αk

N∑

n=1

‖∇en‖2


N−1∑

n=0

e−2αtn‖∇Un+1‖2‖∇en‖2 +ν

6ke−αk

N∑

n=1

‖∇en‖2.(5.41)

Substitute (5.14), (5.15), (5.40) and (5.41) in (5.34). As in the estimate of (5.21),we now apply Gronwall’s Lemma to complete the rest of the proof.

Now a use of Theorems 3.1, 5.1 and (5.30) completes the proof of the followingTheorem.

Theorem 5.2. Under the assumptions of Theorems 3.1 and 5.1, the following holdstrue:

‖u(tn)−Un‖j ≤ Ce−αtn(h2−j + k) j = 0, 1

and

‖(p(tn)− Pn)‖ ≤ Ce−αtn(h+ k).


6. Second Order Backward Difference Scheme

Since the backward Euler method is only first order accurate, we now try toobtain a second order accuracy by employing a second order backward differencescheme. Setting

D(2)t Un =

1

2k(3Un − 4Un−1 +Un−2),(6.1)

we obtain the second order backward difference applied to (3.1) as follows: finda sequence of functions Unn≥1 ∈ Hh and Pnn≥1 ∈ Lh as solutions of thefollowing recursive nonlinear algebraic equations:

(D(2)t Un,φh) + κa(D

(2)t Un,φh) + νa(Un,φh) + b(Un,Un,φh)(6.2)

= (Pn,∇ · φh) ∀φh ∈ Hh,

(∂tU1,φh) + κa(∂tU

1,φh) + νa(U1,φh) + b(U1,U1,φh)

= (P 1,∇ · φh) ∀φh ∈ Hh,

(∇ ·Un, χh) = 0 ∀χh ∈ Lh,

U0 = u0h.

Equivalently, find Unn≥1 ∈ Jh to be the solutions of

(D(2)t Un,φh) + κa(D

(2)t Un,φh) + νa(Un,φh)(6.3)

+ b(Un,Un,φh) = 0 ∀n ≥ 2 ∀φh ∈ Jh,

(∂tU1,φh) + κa(∂tU

1,φh) + νa(U1,φh)

+ b(U1,U1,φh) = 0,

U0 = u0h.

The results of this section are based on the identity which is obtained by a modifi-cation of a similar identity in [1]:

2e2αtn(an, 3an − 4an−1 + an−2) = ‖an‖2 − ‖an−1‖2

+ (1− e2αk)(‖an‖2 + ‖an−1‖2) + ‖δ2an−1‖2(6.4)

+ ‖2an − eαkan−1‖2 − ‖2an−1 − eαkan−2‖2,

where

δ2an−1 = eαkan − 2an−1 + eαkan−2.

Next, we discuss the decay properties for the solution of (6.3).

Lemma 6.1. With 0 ≤ α <νλ1

2(1 + λ1κ), choose k0 small so that for 0 < k ≤ k0

νkλ1κλ1 + 1

+ 1 > e2αk.(6.5)

Then, the discrete solution UN , N ≥ 1 of (6.3) satisfies the following a prioribound:

(‖UN‖2 + κ‖∇UN‖2) + e−2αtN k

N∑

n=1

e2αtn‖∇Un‖2

≤ C(κ, ν, α, λ1)e−2αtN (‖U0‖2 + κ‖∇U0‖2).(6.6)


Proof. Multiply (6.3) by eαtn and substitute φh = Un. Then, using identity (6.4),we obtain

1

4∂t(‖U

n‖2 + κ‖∇Un‖2) + ν‖∇Un‖2 +

(

1− e2αk

4k

)(

‖Un‖2 + κ‖∇Un‖2)

+

(

1− e2αk

4k

)(

‖Un−1‖2 + κ‖∇Un−1‖2)

+1

4k‖δ2Un−1‖2 +

1

4kκ‖δ2∇Un−1‖2

+1

4k

(

(2Un − eαkUn−1)2 − (2Un−1 − eαkUn−2)2)

(6.7)

+κ

4k

(

(2∇Un − eαk∇Un−1)2 − (2∇Un−1 − eαk∇Un−2)2)

= 0.

Note that, the fifth and sixth terms on the left hand side of (6.7) are non-negative.Therefore, we have dropped these terms. Further, observe that

N∑

n=2

(‖Un−1‖2 + κ‖∇Un−1‖2) = (‖U1‖2 + κ‖∇U1‖2)− (‖UN‖2 + κ‖∇UN‖2)

+

N∑

n=2

(‖Un‖2 + κ‖∇Un‖2).(6.8)

Multiplying (6.7) by 4ke−2αk, summing over n = 2 to N , using (2.3) and (6.8), weobtain

‖UN‖2 + κ‖∇UN‖2 + k

(

4νe−2αk − 2(1− e−2αk

k

)(

κ+1

λ1

)

) N∑

n=2

‖∇Un‖2

+‖2e−αkUN − UN−1‖2 + κ‖2e−αk∇UN −∇UN−1‖2 ≤(

‖U1‖2

+κ‖∇U1‖2)

+(

(2e−αkU1 −U0)2 + κ(2e−αk∇U1 −∇U0)2)

.(6.9)

To estimate the first term on the right hand side, we choose n = 1 in (4.9) to obtain

1

2∂t(‖U

1‖2 + κ‖∇U1‖2) +

(


k

)(

κ+1

λ1

)

)

‖∇U1‖2 ≤ 0.(6.10)

Sinceνkλ1

1 + κλ1+ 1 > e2αk, the coefficient of second term becomes positive. There-

fore, we drop this term and obtain

‖U1‖2 + κ‖∇U1‖2 ≤ C(‖U0‖2 + κ‖∇U0‖2).(6.11)

Now, we bound the second term on the right hand side of (6.9) by using CauchySchwarz’s inequality, Young’s inequality and (6.11) as follows:

(2e−αkU1 −U0)2 + κ(2e−αk∇U1 −∇U0)2 ≤ C(κ)(‖U0‖2 + ‖∇U0‖2).(6.12)

Using (6.11), (6.12) and (6.5) in (6.9), we complete the rest of the proof.

Remark. Existence of solution to (6.3) can be proved using Theorem 4.1 andLemma 6.1.

As a consequence of Lemma 6.1, we have the following error estimates.

Theorem 6.1. Assume that 0 ≤ α <νλ1

2(1 + κλ1)and choose k0 ≥ 0 such that for

0 < k ≤ k0,νkλ1

1 + κλ1+ 1 > e2αk.


Let uh(t) be a solution of (3.2) and en = Un − uh(tn), for n = 1, 2, · · · , N . Then,for some positive constant C = C(κ, ν, α, λ1,M), there hold

‖en‖2 + κ‖∇en‖2 + ke−2αtn

n∑

i=2

e2αti‖∇ei‖2 ≤ Ck4e−2αtn(6.13)

and for n = 2, · · · , N ,

‖D2t e

n‖2 + κ‖D2t∇en‖2 ≤ Ck4e−2αtn .(6.14)

Proof. The proof for error analysis is on the similar lines as that of Theorem 5.1.This time the equation in en for n ≥ 2 is

(D(2)t en,φh) + κa(D

(2)t en,φh) + νa(en,φh)(6.15)

= (σn2 ,φh) + κa(σn

2 ,φh) + Λh(φh),

where σn2 and Λh(φh) are defined by

σn2 = un

ht −D(2)t un

h, Λh(φh) = b(unh,u

nh,φh)− b(Un,Un,φh).

Multiply (6.15) by 4keαtn and substitute φh = en. Using identity (6.4), we arriveat

k∂t(‖en‖2 + κ‖∇en‖2) + ‖δ2en−1‖2 + κ‖δ2∇en−1‖2 + 4kν‖∇en‖2(6.16)

+(1− e2αk)(‖en‖2 + κ‖∇en‖2) + (1− e2αk)(‖en−1‖2 + κ‖∇en−1‖2)

+(2en − eαken−1)2 − (2en−1 − eαken−2)2 + κ(2∇en − eαk∇en−1)2

−κ(2∇en−1 − eαk∇en−2)2 = 4k(eαtnσn2 , e

n) + 4k κ a(eαtnσn2 , e

n)

+4k eαtnΛh(en).

Summing (6.16) over n = 2 to N , using (6.8), eo = 0 and dividing by e2αk, weobtain

‖eN‖2 + κ‖∇eN‖2 + e−2αkN∑

n=2

(‖δ2en−1‖2 + κ‖δ2∇en−1‖2) + (2e−αkeN − eN−1)2

+κ(2e−αk∇eN −∇eN−1)2 + k

(

4νe−2αk − 2(1− e−2αk

k

)(

κ+1

λ1

)

) N∑

n=2

‖∇en‖2

≤ ‖e1‖2 + κ‖∇e1‖2 + (2e−αke1 − eo)2 + κ(2e−αk∇e1 −∇eo)2

+4ke−2αkN∑

n=2

(eαtnσn2 , e

n) + 4k κe−2αkN∑

n=2

a(eαtnσn2 , e

n) + 4ke−2αkN∑

n=2

eαtnΛh(en)

≤ C(‖e1‖2 + κ‖∇e1‖2) + I∗1 + I∗2 + I∗3 , say.(6.17)

Now, with the help of Cauchy-Schwarz’s inequality, (2.3) and Young’s inequality,we bound |I∗1 | as:

|I∗1 | ≤ 4ke−2αk(

N∑

n=2

‖eαtnσn2 ‖

2)12 (

N∑

n=2

‖en‖2)12

≤ C(ǫ, λ1)ke−2αk

N∑

n=2

‖eαtnσn2 ‖

2 + ǫke−2αkN∑

n=2

‖∇en‖2.(6.18)


Using ‖unht −D

(2)t un

h‖ ≤ (k)32√2

∫ tntn−2

‖uhttt‖dt ([1]), we note that

‖eαtnσn2 ‖

2 ≤k3

2

∫ tn

tn−2

e2αtn‖uhttt(t)‖2 dt(6.19)

and hence, using (6.19) and Lemma 3.5, we obtain

kN∑

n=2

‖eαtnσn2 ‖

2 ≤k4

2

N∑

n=2

∫ tn

tn−2

e2αtn‖uhttt(t)‖2dt

=k4

2e4αk

N∑

n=2

∫ tn

tn−2

e2αtn−2‖uhttt(t)‖2dt

≤k4

2e4αk

N∑

n=2

∫ tn

tn−2

e2αt‖uhttt(t)‖2dt

≤ k4e4αk∫ tN

0

e2αt‖uhttt(t)‖2dt

≤ C(κ, ν, α, λ1,M)k4e4αke−2αtN .(6.20)

Using (6.20) in (6.18), we arrive at

|I∗1 | ≤ C(κ, ν, α, λ1,M, ǫ)k4 + ǫke−2αkN∑

n=2

‖∇en‖2.(6.21)

Similarly, with the help of Cauchy-Schwarz’s inequality, Young’s inequality andLemma 3.5, we bound

|I∗2 | ≤ C(κ, ν, α, λ1,M, ǫ)k4 + ǫke−2αkN∑

n=2

‖∇en‖2.(6.22)

Once again, a use of (5.18) yields

|I∗3 | ≤ C(ǫ)N∑

n=2

ke−2αke−2αtn‖∇Un‖2‖en‖2(6.23)

+ ǫke−2αkN∑

n=2

‖∇en‖2.

To bound the first term in the right hand side of (6.17), we choose n = 1 in (5.7)and obtain

1

2∂t(

‖e1‖2 + κ‖∇e1‖2)

+

(


k

)(

κ+1

λ1

)

)

‖∇e1‖2(6.24)

= e−αk(eαkσ11 , e

1) + e−αkκa(eαkσ11 , e

1) + e−αkeαkΛh(e1).

On multiplying (6.24) by 2k, using Cauchy-Schwarz’s inequality, (2.3), Young’sinequality appropriately with the estimates (5.20) (for n = 1 and ǫ = ν), we obtain

‖e1‖2 + κ‖∇e1‖2 + 2k

(


k

)(

κ+1

λ1

)

)

‖∇e1‖2(6.25)

≤ 2ke−αk(eαkσ11 , e

1) + 2ke−αkκa(eαkσ11 , e

1) + 2ke−αkeαkΛh(e1)

≤ Ck2e−2αk(

‖eαkσ11‖

2 + κ‖eαk∇σ11‖

2)

+1

2(‖e1‖2 + κ‖∇e1‖2)

+ C(ν)ke−αke−2αk‖∇U1‖2‖e1‖2 + νke−αk‖∇e1‖2


and hence, a use of (5.28) with n = 1 along with (2.3) yields

‖e1‖2 + κ‖∇e1‖2 + k

(

νe−αk − 2(1− e−αk

k

)(

κ+1

λ1

)

)

‖∇e1‖2(6.26)

≤ C(κ, ν, α, λ1,M, ǫ)k4 + C(ν)ke−αke−2αk‖∇U1‖2‖e1‖2.

Using (6.21), (6.22), (6.23) with ǫ = 2ν3 , (6.26), eo = 0 and bounds from Lemma

6.1 in (6.17), we obtain

‖eN‖2 + κ‖∇eN‖2 + e−2αkN∑

n=2

(‖δ2en−1‖2 + κ‖δ2∇en−1‖2) + (2e−αkeN − eN−1)2

+κ(2e−αk∇eN −∇eN−1)2 + 2k

(

νe−2αk −(1− e−2αk

k

)(

κ+1

λ1

)

) N∑

n=2

‖∇en‖2

≤ C(κ, ν, α, λ1,M)k4 + C(ν)

N∑

n=2

ke−2αke−2αtn‖∇Un‖2‖en‖2

+C(ν)ke−αke−2αk‖∇U1‖2‖e1‖2

≤ C(κ, ν, α, λ1,M)k4 + C(ν)

N−1∑

n=0

ke−αke−2αtn‖∇Un‖2‖en‖2

+C(ν)ke−2αke−2αtN‖∇UN‖2‖eN‖2

≤ C(κ, ν, α, λ1,M)k4 + C(ν)

N−1∑

n=0

ke−αke−2αtn‖∇Un‖2‖en‖2

+Cke−2αk(‖eN‖2 + κ‖∇eN‖2).(6.27)

Choose k0, so that (6.5) is satisfied and (1 − Cke−2αk) > 0 for 0 < k ≤ k0. Then,an application of the discrete Gronwall’s Lemma yields

‖eN‖2 + κ‖∇eN‖2 + k

N∑

n=2

‖∇en‖2 ≤ C(κ, ν, α, λ1,M)k4×

exp(kN−1∑

n=0

‖∇Un‖2).(6.28)

The bounds obtained from Lemma 6.1 in (6.28) would lead us to (6.13), that is,

‖eN‖2 + κ‖∇eN‖2 + kN∑

n=2

‖∇en‖2 ≤ C(κ, ν, α, λ1,M)k4(6.29)

and this completes the proof of (6.13) for n ≥ 2. For n = 1, we use (6.26) alongwith the bounds in Lemma 4.1. Then, a choice of k such that (1 − Cke−αk) > 0would lead us to the desired result, that is,

‖e1‖2 + κ‖∇e1‖2 + k‖∇e1‖2 ≤ C(κ, ν, α, λ1,M)k4e−2αk.(6.30)

To arrive at the estimates in (6.14), we choose φh = D(2)t en in (6.15) and obtain

‖D(2)t en‖2 + κ‖∇D

(2)t en‖2 = −νa(en, D

(2)t en) + (σn

2 , D(2)t en)(6.31)

+ κa(σn2 , D

(2)t en) + Λh(D

(2)t en).


It follows from (5.26) that

|Λh(D(2)t en)| ≤ C(κ, ν, α, λ1,M)‖∇en‖‖D

(2)t ∇en‖.(6.32)

Using Cauchy-Schwarz’s inequality, Young’s inequality, (2.3) and (6.32) in (6.31),we arrive at

‖D(2)t en‖2 + κ‖∇D

(2)t en‖2 ≤ C(κ, ν, α, λ1,M)

(

‖∇en‖2 + ‖∇σn2 ‖

2

)

.(6.33)

For the second term on the right hand side of (6.33), we use (6.19), Lemma 3.5 andobtain

‖eαtn∇σn2 ‖

2 ≤k3

2

∫ tn

tn−2

e2αtn‖∇uhttt(t)‖2dt

≤ C(κ, ν, α, λ1,M)k3 e2αtn∫ tn

tn−2

e−2αtdt

≤ C(κ, ν, α, λ1,M)k4e4αk∗

,(6.34)

where k∗ ∈ (0, k). Now with the help of (6.13) and (6.34), (6.33) implies (6.14).This completes the rest of the proof.

Finally, we obtain the error estimates for the pressure Pn. Consider (3.1) at t = tnand subtract it from (6.2) to obtain

(ρn,∇ · φh) = (D2t e

n,φh) + κa(D2t e

n,φh) + νa(en,φh)

− (σn2 ,φh)− κa(σn

2 ,φh)− Λh(φh).

With the help of Cauchy-Schwarz’s inequality, (2.3) and (5.26), we obtain

(ρn,∇ · φh) ≤ C(κ, ν, λ1)(

‖D2t∇en‖+ ‖∇en‖+ ‖∇σn

2 ‖)

‖∇φh‖.(6.35)

A use of the Theorem 6.1 and (6.34) in (6.35) yields

‖ρn‖ ≤ C(κ, ν, α, λ1,M)k2 e−αtn for n ≥ 2.(6.36)

For n = 1, we use estimates obtained from backward Euler method. Substituten = 1 in (5.29) to obtain

(ρ1,∇ · φh) ≤ C(κ, ν, λ1)(

‖∂t∇e1‖+ ‖∇e1‖+ ‖∇σ11‖)

‖∇φh‖.(6.37)

A use of (5.27) with n = 1 in (6.37) yields

(ρ1,∇ · φh) ≤ C(κ, α, ν, λ1,M)(

‖∇e1‖+ ‖∇σ11‖)

‖∇φh‖.(6.38)

Using bounds obtained from (5.28) (for n = 1) and (6.30) in (6.38), we arrive at

‖ρ1‖ ≤ C(κ, α, ν, λ1,M)k e−αt1 .(6.39)

Theorem 6.2. Under the assumption of Theorems 3.1, 5.1 and 6.1, the followingholds true:

‖u(tn)−Un‖j ≤ C(h2−j + k2)e−αtn j = 0, 1

and

‖(p(tn)− Pn)‖ ≤ Ce−αtn(h+ k2−γ),

where

γ =

0 if n ≥ 2;

1 if n = 1.

Proof of Theorem 6.2. A use of Theorems 3.1, 6.1, (6.30), (6.36) and (6.39)would complete the proof.


7. Numerical Experiments

In this section, we provide a few computational results to support our theoreticalestimates for the equations of motion arising in the Kelvin-Voigt fluid (1.1)-(1.3).In example 1, for space discretization, P2-P0 mixed finite element space (see [5]) isused: the velocity space consists of continuous piecewise polynomials of degree lessthan or equal to 2 and the pressure space consists of piecewise constants, that is,we consider the finite dimensional subspaces Vh andWh of H1

0 and L2 respectively,as:

Vh = v ∈ (H10 (Ω))

2 ∩(

C(Ω))2

: v|K ∈ (P2(K))2,K ∈ τh,

Wh = q ∈ L2(Ω) : q|K ∈ P0(K),K ∈ τh,

where τh denotes the triangulation of the domain Ω. Below, we discuss the fullydiscrete finite element formulation of (1.1)-(1.3) using backward Euler method andsecond order backward difference scheme.Fully discrete finite element approximation: In this scheme, we discuss thediscretization of the time variable by replacing the time derivative by differencequotient. Let ∆t be the time step and Un be the approximation of u(t) in Vh att = tn = n∆t.The backward Euler approximation to (3.1) can be stated as: given Un−1, find thepair (Un, Pn) satisfying:

(Un,vh) + (κ+ ν∆t) a(Un,vh) + ∆t c(Un,Un,vh) + ∆t b(vh, Pn)(7.1)

= (Un−1,vh) + κa(Un−1,vh) + ∆t (f(tn),vh) ∀vh ∈ Vh,

b(Un, wh) = 0 ∀wh ∈Wh.

Similarly, the second order backward difference approximation to (3.1) is as follows:given Un−2 and Un−1, find the pair (Un, Pn) satisfying:

(3Un,vh) + (κ+ 2ν∆t) a(Un,vh) + 2∆t c(Un,Un,vh)(7.2)

+2∆t b(vh, Pn) = 4(Un−1,vh) + 4 κa(Un−1,vh)− (Un−2,vh)

−κ a(Un−2,vh) + ∆t (f(tn),vh) ∀vh ∈ Vh,

b(Un, wh) = 0 ∀wh ∈ Wh.

Now, we approximate the velocity and pressure by

Un =

ng∑

j=1

(

unxj

unyj

)

φu

j (x), Pn =

ne∑

j=1

pnj φpj (x),(7.3)

where φu

j (x) and φpj (x) form bases for Vh and Wh respectively with cardinality ng

and ne, respectively. Here, unxj and uny

j represent the x and y component of theapproximate velocity field, respectively, at time t = tn.Using (7.3), the basis functions for Vh and Wh in (7.1) (respectively (7.2)), weobtain nonlinear systems which are solved using Newton’s method.Example 1: In this example, we choose the right hand side function f in such away that the exact solution (u, p) = ((u1, u2), p) is

u1 = 10e−tx2(x − 1)2y(y − 1)(2y − 1), u2 = −10e−ty2(y − 1)2x(x − 1)(2x − 1),p = e−ty.

We choose ν = 1, κ = 10−2 with Ω = (0, 1) × (0, 1) and time t = [0, 1]. Here,Ω is subdivided into triangles with mesh size h. The theoretical analysis provides


a convergence rate of O(h2) in L2-norm, of O(h) in H1-norm for velocity and ofO(h) in L2-norm for pressure. Table 1 gives the numerical errors and convergencerates obtained on successively refined meshes for the first order backward Eulermethod and Table 2 contains the errors and convergence rates of the second or-der two step backward difference method. These results agree with the optimaltheoretical convergence rates obtained in Theorem 5.2 and 6.2.

Table 1. Errors and Convergence rates for backward Eulermethod with k = O(h2).

h ‖u(tn)−Un‖L2 Rate ‖u(tn)−U

n‖H1 Rate ‖p(tn)− Pn‖ Rate

1/2 0.0045761 0.056318 0.136225

1/4 0.0013220 1.791328 0.024171 1.220311 0.072946 0.901096

1/8 0.0003651 1.856036 0.010997 1.136107 0.037920 0.943847

1/16 0.0000970 1.911519 0.005371 1.033759 0.019201 0.981790

Table 2. Errors and Convergence rates for backward differencescheme with k = O(h).



1/2 0.0047034 0.043623 0.135666

1/4 0.0013240 1.828747 0.019412 1.168111 0.073007 0.893959

1/8 0.0003653 1.857587 0.009233 1.072039 0.037925 0.944881

1/16 0.0000970 1.912022 0.004602 1.004511 0.019201 0.981941

Remark 2. Note that, under extra regularity assumptions on the exact solutionpair, one can obtain better rates of convergence by using higher order finite elementspace.

We illustrate this in example 2, by choosing an appropriate right hand side functionf . Example 2: In this example, we choose the right hand side function f in sucha way that the exact solution (u, p) = ((u1, u2), p) is:

u1 = te−t2sin2(3πx) sin(6πy), u2 = −te−t2sin2(3πy) sin(6πx),(7.4)

p = te−t sin(2πx) sin(2πy).

We assume that the viscosity of the fluid(ν) is 10−2 and the retardation κ is 10−4

with Ω = (0, 1)×(0, 1) and time t = [0, 1]. Here again, Ω is subdivided into triangleswith mesh size h. For the problem defined in example 2, we have conducted numer-ical experiments using P2-P1 mixed finite element spaces for space discretization,that is, if we choose

Vh = v ∈ (H10 (Ω))

2 ∩(

C(Ω))2

: v|K ∈ (P2(K))2,K ∈ τh,

Wh = q ∈ L2(Ω) ∩C(Ω) : q|K ∈ P1(K),K ∈ τh,

we obtain ‖u(tn) −Un‖j ≤ O(h3−j), j = 0, 1 and ‖(p(tn) − Pn)‖ ≤ O(h2). Sincethe solution (7.4) has extra regularity, we obtained better order of convergencefor velocity and pressure as expected [5]. In Tables 3 and 4, we have shown theconvergence rates for backward Euler method and backward difference scheme re-spectively for L2 and H1-norms in velocity and L2-norm in pressure. In case, wechoose k = O(h3/2) for backward Euler method, we observe that convergence rate


in comparison with that of the backward difference scheme is lower. Table 5 rep-resents the comparison between the errors obtained from backward Euler methodand backward difference scheme with k = O(h3/2).

Table 3. Errors and Convergence rates for backward Eulermethod with k = O(h3).



1/4 0.480585 13.147142 0.088453

1/8 0.085185 2.496114 4.135230 1.668709 0.015389 2.522972

1/16 0.007371 3.530504 0.849642 2.283040 0.002566 2.584331

1/32 0.000709 3.377407 0.163177 2.380417 0.000610 2.072467

Table 4. Errors and Convergence rates for backward differencescheme with k = O(h3/2).


n‖H1 Rate ‖p(tn) − Pn‖ Rate

1/4 0.466215 12.744857 0.085101

1/8 0.083276 2.485017 4.111723 1.63210 0.015361 2.469883

1/16 0.007224 3.526926 0.850460 2.273426 0.002504 2.616902

1/32 0.000609 3.568176 0.163161 2.381940 0.000596 2.068971

Table 5. Comparison of errors with k = O(h3/2) between the two schemes.

h BE velocity Bd velocity BE pressure Bd pressurein L

2-norm in L2-norm in L

2-norm in L2-norm

1/4 0.514184 0.466215 0.102280 0.085101

1/8 0.084014 0.083276 0.015499 0.015361

1/16 0.008783 0.007224 0.003010 0.002504

1/32 0.001991 0.000609 0.000880 0.000596

Here BE = Backward Euler, Bd = Backward difference.

Acknowledgments

The authors acknowledge the financial support provided by the DST-CNPq Indo-Brazil Project No. DST/INT/Brazil/RPO-05/2007 (Grant no. 490795/2007-2).Further,they thank the referees for their valuable comments.

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Department of Mathematics, Industrial Mathematics Group, Indian Institute of TechnologyBombay, Powai, Mumbai-400076, India.

E-mail : [email protected], [email protected] and [email protected]

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